A certain circle can be represented by the following equation. $x^2-16x+y^2-36=0$ What is the center of this circle ? $($
Explanation: The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2-16x+y^2-36&=0\\\\ x^2-16x+y^2&=36\\\\ (x^2-16x)+y^2&=36 \text{(grouping terms)}\\\\ (x^2-16x{+64})+y^2&=36{+64}\end{aligned}$ Notice that we must add ${64}$ on the right side of the equation, since we added it to the left side of the equation. [How did we get 64?] Writing the equation in standard form $\begin{aligned}(x^2-16x{+64})+y^2&=36{+64}\\\\ (x-8)^2+(y-0)^2&=100\\\\ (x-8)^2+(y-0)^2&=10^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(8,0)$ and has a radius of $10$ units. Summary The circle is centered at $(8,0)$. The circle has a radius of $10$ units.